Proposition 18

The sequence of $A[X]$-homomorphisms

is exact.

As $\phi(1 \otimes x) = x$ for all $x \in M$, clearly $\phi$ is surjective; on the other hand,

in other words, $\phi \circ X = u \circ \phi = \phi \circ \bar{u}$ by (46); this proves that $\phi \circ \psi = 0$. It remains to verify that $\operatorname{Ker} \phi \subset \operatorname{Im} \psi$. For this note that, since the monomials $X^k$ ($k \geq 0$) form a basis of the $A$-module $A[X]$, every element $z \in M[X]$ can be written uniquely in the form $z = \sum_k X^k \otimes x_k$, where $(x_k)$ is a family of elements of $M$, of finite support. If $z \in \operatorname{Ker} \phi$, then $\phi(z) = \sum_k u^k(x_k) = 0$ and we can write But as the $A[X]$-endomorphisms $X$ and $\bar{u}$ of $M[X]$ are permutable, then

which proves that there exists a $y \in M[X]$ such that $z = \psi(y)$.

Now let $M'$ be another $A$-module and $u'$ an endomorphism of $M'$; let $M'_{u'}$, $\phi'$, $\bar{u}'$, $\psi'$ be the module and mapping obtained from $M'$ and $u'$ as $M_u$, $\phi$, $\bar{u}$, $\psi$ are obtained from $M$ and $u$. Then: